Brain Teasers & Riddles

[Michael]

Why is an island like the letter "t" ?

[Margot]

Why is an island like the letter "t" ?

Both are in the middle of water?

[The Drunk Guy]

Both are in the middle of water?
Can't say that's wrong.

Have I ever expressed my earnest, life-long love of beer, ales, and lagers?:drink:

[Michael]

Both are in the middle of water?

Correct! :)

Your turn to post a riddle/puzzle next.

[Margot]

OK. Hope this one hasn't already been done.

What's the beginning of eternity, the end of time and space, the start of every end, and the end of every race?

[Michael]

OK. Hope this one hasn't already been done.

What's the beginning of eternity, the end of time and space, the start of every end, and the end of every race?
The letter "e". :)

[Margot]

The letter "e". :)

Yup. That one works much better when spoken.

[Michael]

Yup. That one works much better when spoken.
Indeed it does! :D

Next one...

What is it that you can keep after giving it to someone else?

[Dominick]

Indeed it does! :D

Next one...

What is it that you can keep after giving it to someone else?
Syphilis :angel:

[Michael]

Syphilis :angel:
I gotta give that one to you! :rofl:

That's not the answer I was looking for, but that answer certainly answers the riddle well! :D

[Dominick]

I gotta give that one to you! :rofl:
:eek:
In the context of the previous posts, that's a disturbing answer. :erm:

[Michael]

:eek:
In the context of the previous posts, that's a disturbing answer. :erm:
:eek:

Good point! I'm saying I'm acknowledging your answer is sufficiently correct. :)

And I don't have any syphilis to give to you anyway! ;)

[Dominick]

Glad we got that out of the way :D

Anyway, what room has no windows or doors ?

[Evangeline]

Glad we got that out of the way :D

Anyway, what room has no windows or doors ?

A chat room?

[Dominick]

A chat room?
Hmmm. Not the answer I was looking for but I'll give it to you.

Your turn to come up with a riddle !

PS the original answer was 'mushroom'. :o

[Evangeline]

I dig out tiny caves, and store gold and silver in them. I also build bridges of silver and make crowns of gold. They are the smallest you could imagine. Sooner or later everybody needs my help, yet many people are afraid to let me help them. Who am I?

[Greendruid]

I dig out tiny caves, and store gold and silver in them. I also build bridges of silver and make crowns of gold. They are the smallest you could imagine. Sooner or later everybody needs my help, yet many people are afraid to let me help them. Who am I?

You are a dentist!?!?

[Evangeline]

You are a dentist!?!?

Yup! You got it.

[Michael]

Since Greendruid is apparently catching the Dominick-disease...

* * *

Give a word that contains all SIX vowels in order (including Y as a vowel).

[The Drunk Girl]

Since Greendruid is apparently catching the Dominick-disease...

* * *

Give a word that contains all SIX vowels in order (including Y as a vowel).

adventitiously

[Michael]

adventitiously
That's all six vowels in a row, but is that a word? :ummm:

I don't think so...

[Dominick]

That's all six vowels in a row, but is that a word? :ummm:

I don't think so...
It's in the OED.

[Michael]

It's in the OED.
Well then, I stand corrected. Does it come with a definition too?

[Dominick]

Well then, I stand corrected. Does it come with a definition too?
Yes .

[Dominick]

Ok, ok, you might want to actually know that definition :)


adventitious adj. 1 happening according to chance 2 coming from outside, not native 3 Biology [...]
- Derivatives adventitiously adv.

[Michael]

Ok, ok, you might want to actually know that definition :)
Interesting and yes. The word seems superfluous though. :)

Here's another challenge...

How about a five letter word with four vowels in it? (English only, not counting "y" as a vowel this time).

[Dominick]

Interesting and yes. The word seems superfluous though. :)

Here's another challenge...

How about a five letter word with four vowels in it? (English only, not counting "y" as a vowel this time).
aerie

And I didn't even had to read the entire OED. :angel:

Or do it need to be different vowels?

Edit : just remembered where I know it from : Jefferson Airplane !

[Michael]

aerie

And I didn't even had to read the entire OED. :angel:

That one works!

(even if you missed that one, audio comes up next in the dictionary)

[Michael]

Okay... here's another one.

A woman was being chased through the jungle by a tribe of hunters. She had in her possession two solid gold ingots which weighed 9 pounds each. She came to a rope bridge whcih she needed to cross, but she knew could only support 125 pounds in weight. She herself weighed 110 pounds. She did not have time to carry the ingots across one at a time and the distance was too great to throw them. For the sake of the puzzle, we'll assume the woman is naked too so doesn't have any clothes to discard. The hunters were close behind.

How can she escape across the bridge with both gold ingots?

[dilettante]

Okay... here's another one.

A woman was being chased through the jungle by a tribe of hunters. She had in her possession two solid gold ingots which weighed 9 pounds each. She came to a rope bridge whcih she needed to cross, but she knew could only support 125 pounds in weight. She herself weighed 110 pounds. She did not have time to carry the ingots across one at a time and the distance was too great to throw them. For the sake of the puzzle, we'll assume the woman is naked too so doesn't have any clothes to discard. The hunters were close behind.

How can she escape across the bridge with both gold ingots?

Can we get a hint? :)

[drgoodtrips]

Okay... here's another one.

A woman was being chased through the jungle by a tribe of hunters. She had in her possession two solid gold ingots which weighed 9 pounds each. She came to a rope bridge whcih she needed to cross, but she knew could only support 125 pounds in weight. She herself weighed 110 pounds. She did not have time to carry the ingots across one at a time and the distance was too great to throw them. For the sake of the puzzle, we'll assume the woman is naked too so doesn't have any clothes to discard. The hunters were close behind.

How can she escape across the bridge with both gold ingots?

I suppose she could juggle them while crossing.

[Michael]

I suppose she could juggle them while crossing.
Bingo! Give drgoodtrips a cookie! :)

[drgoodtrips]

All right, here's one:

A (very) simple farmer had a large quantity of eggs to sell to the nearest farmer's market.

When the farmer's market purchaser asked him how many eggs he wanted to sell, he said, "son, I can't count past 100 But, I do know that...

If you group the eggs in 2's there will be one egg left.
If you group the eggs in 3's there will be one egg left.
If you group the eggs in 4's there will be one egg left.
If you group the eggs in 5's there will be one egg left.
If you group the eggs in 6's there will be one egg left.
If you group the eggs in 7's there will be one egg left.
If you group the eggs in 8's there will be one egg left.
If you group the eggs in 9's there will be one egg left.
If you group the eggs in 10's there will be one egg left.

Finally. If you group the eggs in 11's there will be no eggs left!"

How many eggs does the farmer have?

[The Drunk Girl]

All right, here's one:

Is this saying there is exactly only one egg left?

[Michael]

The farmer has 3,628,801 eggs! :)

x = 2*3*4*5*6*7*8*9*10+1

[drgoodtrips]

Is this saying there is exactly only one egg left?

I'm not sure I know what you mean. Are you giving an answer of "1"?

[drgoodtrips]

The farmer has 3,628,801 eggs! :)

x = 2*3*4*5*6*7*8*9*10+1

No, that would't work. 3,628,801 cannot be evenly divided by 11 (though your answer does satisfy all but one of the constraints).

Edit: My apologies - that answer does work (I must have mistyped initially into the calculator). There is a much smaller answer, but yours is correct, since it satisfies all constraints.

[Michael]

No, that would't work. 3,628,801 cannot be evenly divided by 11 (though your answer does satisfy all but one of the constraints).

Edit: My apologies - that answer does work (I must have mistyped initially into the calculator). There is a much smaller answer, but yours is correct, since it satisfies all constraints.
Well then, perhaps we should leave this up for a day or two to see if any math whiz kids want to do it - I'd suspect one might factor out the two... (no... that doesn't work) :ummm:

[Dominick]

27,731 ?

[drgoodtrips]

27,731 ?

No. Modulo arithmetic there doesn't work out (though you are pretty close, numerically - I can't speak to the thought process).

[drgoodtrips]

Well then, perhaps we should leave this up for a day or two to see if any math whiz kids want to do it - I'd suspect one might factor out the two... (no... that doesn't work) :ummm:

You did the first half right, and have the right idea for the second half yourself... :cool:

[Dominick]

No. Modulo arithmetic there doesn't work out (though you are pretty close, numerically - I can't speak to the thought process).
Damn, I missed a math question. My reputation is gone to hell. :eek:

Edit : <slaps head> of course, it's 25,201

[drgoodtrips]

Yes, that's it :D

[wphelan]

Yes, that's it :D

Can the mathematically disinclined get a solution for that?

[Dominick]

Can the mathematically disinclined get a solution for that?
First you consider just this:

If you group the eggs in 2's there will be one egg left.
If you group the eggs in 3's there will be one egg left.
If you group the eggs in 4's there will be one egg left.
If you group the eggs in 5's there will be one egg left.
If you group the eggs in 6's there will be one egg left.
If you group the eggs in 7's there will be one egg left.
If you group the eggs in 8's there will be one egg left.
If you group the eggs in 9's there will be one egg left.
If you group the eggs in 10's there will be one egg left.

Since the number we seek always leaves 1 when we divide it by all numbers from 2 to 10 we know that the (number-1) would always leave zero, which means the number minus one is divisible by every number from 2 to 10. To find (number-1) we therefore need the least common multiple (http://en.wikipedia.org/wiki/Least_common_multiple) of all numbers from 2 to ten.
Using any of the techniques in the wiki link you'll find that the LCM of the numbers 2 to 10 is 2,520. But our result is (number-1) so the number is 2,521. Or at least it would be without the last condition:

Finally. If you group the eggs in 11's there will be no eggs left!"
Now, we need to track back a little and find a multiple of (number-1) that's divisible by 11 when you add one to it. (This is where I went wrong in my first guess - I erroneously used multiples of number instead of multiples of (number-1)). And that multiple turns out to be 25,200 since adding 1 gives 25,201 being divisible by 11.

Note that it's much more difficult to explain it than to actually do it :)

[Michael]

This game works on the honor system. Google and/or other search engines are not allowed to be used for finding the answers.


To find (number-1) we therefore need the least common multiple (http://en.wikipedia.org/wiki/Least_common_multiple) of all numbers from 2 to ten.

I don't think one ought to be using wiki (or a book) for looking up info here. Brain teasers aren't research questions.

[Dominick]

I don't think one ought to be using wiki (or a book) for looking up info here. Brain teasers aren't research questions.
It's for the explanation. I don't need to use wiki to find an LCM, geez. Heck, I don't even need a ballpoint. Lol !

[drgoodtrips]

First you consider just this:

Since the number we seek always leaves 1 when we divide it by all numbers from 2 to 10 we know that the (number-1) would always leave zero, which means the number minus one is divisible by every number from 2 to 10. To find (number-1) we therefore need the least common multiple (http://en.wikipedia.org/wiki/Least_common_multiple) of all numbers from 2 to ten.
Using any of the techniques in the wiki link you'll find that the LCM of the numbers 2 to 10 is 2,520. But our result is (number-1) so the number is 2,521. Or at least it would be without the last condition:

Now, we need to track back a little and find a multiple of (number-1) that's divisible by 11 when you add one to it. (This is where I went wrong in my first guess - I erroneously used multiples of number instead of multiples of (number-1)). And that multiple turns out to be 25,200 since adding 1 gives 25,201 being divisible by 11.

Note that it's much more difficult to explain it than to actually do it :)

You know, there are certain things that "just happen" when I do math. LCM or coming up with prime factors is definitely one of them. I remember trying to explain it to someone recently who wanted some help with a math problem. I explained what prime factors of a number were, but was at a loss when asked how to find them (i.e. the actual algorithm). I replied sort of sheepishly, "I don't know."

Interestingly, if asked to write a program to do it, that'd be a snap. But, it's also certainly not how my brain works (at least not until I complete my life's master work of installing a fast processor in my brain).

/shrug/

[Michael]

Damn, I missed a math question. My reputation is gone to hell. :eek:

Edit : <slaps head> of course, it's 25,201

I'll split the 'credit' with you here. Since my answer was indeed correct and first, but you've got the 'more correct' answer drgoodtrips was looking for. :)

And being the magnanimous fellow that I am, I shall offer you the honor of setting the next challenge. :rofl:

[dilettante]

Okay... here's another one.

A woman was being chased through the jungle by a tribe of hunters. She had in her possession two solid gold ingots which weighed 9 pounds each. She came to a rope bridge whcih she needed to cross, but she knew could only support 125 pounds in weight. She herself weighed 110 pounds. She did not have time to carry the ingots across one at a time and the distance was too great to throw them. For the sake of the puzzle, we'll assume the woman is naked too so doesn't have any clothes to discard. The hunters were close behind.

How can she escape across the bridge with both gold ingots?

I suppose she could juggle them while crossing.

Bingo! Give drgoodtrips a cookie! :)

OK, this puzzle has been bothering me for days now, and after some furious calculations this afternoon, I'm putting my nerdiness on display and challenging this answer. I don't think juggling helps in the least; in fact, I think in a real world scenario it would only make the bridge more likely to break.

The problem is that in throwing one idol up into the air, the women exerts a downward force on the bridge equal to the upward force she exerts on the idol. If she were to juggle the idols perfectly (meaning that she used the minimum force necessary to keep one of them in the air at all times) she would be exerting a force on the bridge precisely equal to that she would exert simply holding the two idols the in her hands. And, as we know, the bridge cannot support that much force.

Numerically, if you work it all out, assuming that one idol is in the air, the women can exert up to 66.64 N up on the idol she is holding. That's enough to give the idol an upward acceleration of 6.53 m/s^2, which is less than the 9.8 m/s^2 of downward acceleration that gravity will exert on the idol after it leaves her hand. Therefore, gravity will return the first idol she tosses back to her hand faster than she can catch and throw (i.e. re-accelerate) the second idol. She's doomed.

[drgoodtrips]

OK, this puzzle has been bothering me for days now, and after some furious calculations this afternoon, I'm putting my nerdiness on display and challenging this answer. I don't think juggling helps in the least; in fact, I think in a real world scenario it would only make the bridge more likely to break.

The problem is that in throwing one idol up into the air, the women exerts a downward force on the bridge equal to the upward force she exerts on the idol. If she were to juggle the idols perfectly (meaning that she used the minimum force necessary to keep one of them in the air at all times) she would be exerting a force on the bridge precisely equal to that she would exert simply holding the two idols the in her hands. And, as we know, the bridge cannot support that much force.

Numerically, if you work it all out, assuming that one idol is in the air, the women can exert up to 66.64 N up on the idol she is holding. That's enough to give the idol an upward acceleration of 6.53 m/s^2, which is less than the 9.8 m/s^2 of downward acceleration that gravity will exert on the idol after it leaves her hand. Therefore, gravity will return the first idol she tosses back to her hand faster than she can catch and throw (i.e. re-accelerate) the second idol. She's doomed.

I'm not sure how you reach that conclusion.

Also, this is not a zero sum game in force with the bridge. That is, she could, in principle, drop the idols, catch them at a lower height, and then lift them up again without exerting any force against the bridge whatsoever. The only energy would come from her burning calories.

So, theoretically, she could drop one, accelerate her hand downward at a higher rate than the acceleration of gravity, catch it, and accelerate her hand with the idol back up to drop again. If she staggered this with both hands, there would be no issue at all (the difficulty of this task from a manual dexterity standpoint notwithstanding).

[dilettante]

I'm not sure how you reach that conclusion.

Also, this is not a zero sum game in force with the bridge. That is, she could, in principle, drop the idols, catch them at a lower height, and then lift them up again without exerting any force against the bridge whatsoever. The only energy would come from her burning calories.

So, theoretically, she could drop one, accelerate her hand downward at a higher rate than the acceleration of gravity, catch it, and accelerate her hand with the idol back up to drop again. If she staggered this with both hands, there would be no issue at all (the difficulty of this task from a manual dexterity standpoint notwithstanding).

I'm fairly certain she'd be required to exert a force against the bridge in order to lift or toss the idol. To push up on the idol she must, in effect, push down on the bridge.
Or, to put it another way, when she pushes up on the idol, it pushes back on her (with an equal an opposite force). This force is translated through her to the bridge.

One could also imagine that the women is floating in space. If she throws an idol "up" she is propelled "down" (away from the throw) because the of the force exerted by the idol as she throws it. The women in the scenario doesn't move "down" when she tosses the idol because she's standing on the bridge and the bridge absorbs the additional force (translating it to the ground).
It's the same principle as firing a gun. The women is "firing" the idol up into the air and therefore "recoils" (however slightly) back against the bridge in proportion to the velocity of the toss. The burned calories provide energy for the toss, but do nothing to compensate for the recoil.

[dilettante]

It wasn't actually my intention to kill this thread by challenging a question/answer. :o
So I'll just throw out another brain teaser:



What's special about this sentence?

"I do not know where family doctors acquired illegibly perplexing handwriting, nevertheless extraordinary pharmaceutical intellectuality counterbalancing indecipherability transcendentalizes intercommunications incomprehensibleness."

[Margot]

It wasn't actually my intention to kill this thread by challenging a question/answer. :o
So I'll just throw out another brain teaser:



What's special about this sentence?

"I do not know where family doctors acquired illegibly perplexing handwriting, nevertheless extraordinary pharmaceutical intellectuality counterbalancing indecipherability transcendentalizes intercommunications incomprehensibleness."

Each word is one letter longer than the previous

[dilettante]

Each word is one letter longer than the previous

BING!

Your turn.

[drgoodtrips]

I'm fairly certain she'd be required to exert a force against the bridge in order to lift or toss the idol. To push up on the idol she must, in effect, push down on the bridge.
Or, to put it another way, when she pushes up on the idol, it pushes back on her (with an equal an opposite force). This force is translated through her to the bridge.

Right, but to drop the idol, she must do nothing with the bridge. And, to catch it, she can space the catch out over some amount of time. (F=ma=m*(vf-vi)/(tf-ti)). Spacing out the timing reduces force.

[dilettante]

Right, but to drop the idol, she must do nothing with the bridge.

True, but in order to give it an upward acceleration, whether lifting or tossing it, she'll have to exert a force on the bridge equal to the force she imparts to the idol.


And, to catch it, she can space the catch out over some amount of time. (F=ma=m*(vf-vi)/(tf-ti)). Spacing out the timing reduces force.

Also true, but here's the catch: the longer it takes her to catch one idol, the longer the other idol has to stay in the air. The length of time the idol stays in the air is related to the height it obtains before falling, which is related to the amount of upward acceleration it received, which is, in turn, related to the amount of force she exerted on it when lifting/tossing it and thus to the amount of force exerted on the bridge.

Consider just tossing one idol up and down in such a way that it spends exactly 1/2 its time in the air. While in the air its experiencing an acceleration of 9.8 m/s^2 downward due to gravity. Therefore, gravity is exerting a force of 9.8M downward on the idol (where M is the mass of the idol in kg).
In order for the idol to spend exactly the same amount of time in the woman's hand, it must experience an acceleration exactly opposite to that imparted by gravity when it was in the air. Therefore, the woman must accelerate the idol at 9.8 m/s^2 upward while catching and tossing it. That would require exerting a force of 9.8M upward on the idol and therefore an equal force downward on the bridge.
HOWEVER, gravity is still in play when the the idol is in the woman's hand and still exerting a force of 9.8M downward on the idol. Therefore, the women must exert enough force BOTH to counter-act gravity (and thus stop the downward acceleration) AND to accelerate the idol upward at 9.8 m/s^2. She must therefore exert an upward force of 2*9.8M (or 19.6M) on the idol while catching/tossing it if the idol is to spend exactly half its time in the air.
Since it takes an upward force of 9.8M (just enough to counter-act gravity) the required force of 2*9.8M is the same as just holding two idols still.



EDIT: Oh! Or better yet:

Gravity is exerting a force of 9.8M on the idols the whole time, for a total constant force of 2*9.8M downward. Unless the women is exerting, on average, an equal force upward then the idols MUST move down over time. Every newton of force she exerts upward on the idol ends up being exerted downward on the bridge, and the bridge cannot support 2*9.8M of force in addition to the women. She's doomed.

[dilettante]

Let's see if we can bring this thread back to life:

What would the sixth word in this series be?

Bugles, Unrest, Grotto, Letter, Esteem, ???

[Michael]

Let's see if we can bring this thread back to life:

What would the sixth word in this series be?

Bugles, Unrest, Grotto, Letter, Esteem, ???

Symbol! :)

[dilettante]

Symbol! :)

Hmmm. That wasn't the answer I had in mind. Why Symbol?

[Michael]

Hmmm. That wasn't the answer I had in mind. Why Symbol?

Six letter word starting with "s".

Each of the other five words in the series are six letter words starting with each letter of the first word in sequence (bugles). Thus, the sixth word must begin with "S" and have six letters.

[dilettante]

Six letter word starting with "s".

Each of the other five words in the series are six letter words starting with each letter of the first word in sequence (bugles). Thus, the sixth word must begin with "S" and have six letters.

Ah. Well, yes, all of that is true. However, if you extend that thinking further you'll find that only one specific six letter word beginning with 'S' actually follows this sequence.

[Americano]

It's a 6x6 grid but I'm not going to pop the answer as I don't want to have to come up with the next riddle.

[Michael]

It's a 6x6 grid but I'm not going to pop the answer as I don't want to have to come up with the next riddle.

If you've got the answer, go for it.

If you don't have the next riddle/puzzle, I will put one up.

[wphelan]

Storms

[dilettante]

Storms

Correct.
As a 6x6 grid the words are the same vertically and horizontally.

B U G L E S
U N R E S T
G R O T T O
L E T T E R
E S T E E M
S T O R M S

[Michael]

Okay, here's an easy one...

What word can be written forward, backward or upside down, and can still be read from left to right?

[Greendruid]

NOON of course!

[Michael]

NOON of course!

Right you are. I did say it was an easy one - I was just hoping someone besides you would get it! ;)

[Michael]

24 hour rule...

Here's another puzzle...

There once was an evil wizard. He took 3 woman from their homes and turned them into rose bushes that looked exactly alike. He put them in his garden. One of the woman had a husband and children and begged the wizard to let her see them. He agreed. At night, the wizard brought the woman to her house. In the morning he came and took her back to his garden.

The husband decided to go rescue her. So he snuck into the wizard's garden. He looked and looked at the 3 identical rose bushes trying to figure out which could be his wife. Suddenly, he knew the answer and he took his wife home. How did he know which rose bush was his wife?

[drgoodtrips]

I'm guessing that he picked the rose bush that had recently been transplanted... :D

[dilettante]

I'm guessing that he picked the rose bush that had recently been transplanted... :D

Specifically, I'd guess he picked the one without any dew on it.

[wphelan]

Did the husband know that the wizard took the wife to visit the house? It doesn't say the husband was aware of any of that. :shrug:

[Michael]

Did the husband know that the wizard took the wife to visit the house? It doesn't say the husband was aware of any of that. :shrug:
Given that the husband is presumed to live at that house and his presence might be presumed to be the object of the wife's desire to visit there, I'd say that it is a reasonable assumption that the husband was aware of the visit.

[Michael]

Specifically, I'd guess he picked the one without any dew on it.
Correct.

[wphelan]

Given that the husband is presumed to live at that house and his presence might be presumed to be the object of the wife's desire to visit there, I'd say that it is a reasonable assumption that the husband was aware of the visit.

Ah. For some reason I pictured the wizard holding a plant in front of a window while they watched in a 'ghost of Christmas present' style, and leaving it/her outside until morning.

[Michael]

Ah. For some reason I pictured the wizard holding a plant in front of a window while they watched in a 'ghost of Christmas present' style, and leaving it/her outside until morning.

Yeah well, I loved this puzzle because it reminded me of a Monty Python skit...

<cue English accent>

Mrs. Cranston is hiding behind one of these three bushes (they show three small identical shrubs in a field).

Which shrub is she hiding behind?

(show guy pushing detonator and the first bush explodes like dynamite)

No, Mrs. Cranston is not hiding behind bush #1.

(show guy pushing detonator and the second bush explodes like dynamite with the sound of a lady screaming and a fake body goes flying)

Yes, Mrs. Cranston was hiding behind bush #2.

:rofl:

[dilettante]

There are three boxes, all labeled incorrectly. One contains only nails, one contains only screws, and one contains a mixture of nails and screws. At present, they are labeled as follows:

[box 1] nails
[box 2] screws
[box 3] nails and screws

You are tasked with giving the boxes their correct labels. To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.

Can this be done? If so, how?

[drgoodtrips]

I'd pick one from the box labelled "nails and screws". If you pick a nail, remove the "nails" label and put it on that box. Likewise for a screw. Then, take the label you didn't remove ("nails" or "screws") and move it to the box with the newly vacated label. Finally, take the "nails and screws" label and put it on the box that is now missing a label, and you're done.

[dilettante]

I'd pick one from the box labelled "nails and screws". If you pick a nail, remove the "nails" label and put it on that box. Likewise for a screw. Then, take the label you didn't remove ("nails" or "screws") and move it to the box with the newly vacated label. Finally, take the "nails and screws" label and put it on the box that is now missing a label, and you're done.

Brilliant as ever, good doctor. Your turn.

[drgoodtrips]

Hmm... drawing a blank right now. I'll see if I can think of one before the end of the work day, though if anyone else has one at the ready, I've no objection to you jumping in.

[Michael]

Riddle me this!

Whoever makes it, tells it not. Whoever takes it, knows it not. And whoever knows it wants it not.

What is it?

[Zarquon]

Counterfeit money.

[Michael]

Counterfeit money.

Correct!

Now it is your turn to post a new riddle or puzzle. :)

* * *

Btw, for the noobs, our informal "24 hour rule" here means that if you don't have one or can't find or think of one, then anyone else can just jump in and post the next riddle or puzzle, 24 hours after this post!

We also keep a running scoreboard on the first page of this thread. :)

[Zarquon]

Okay, an easy one:
make a sentence containing all the letters of the English alphabet at least once.

[Michael]

Okay, an easy one:
make a sentence containing all the letters of the English alphabet at least once.

I remember this from high school typing class! :D

"the quick brown fox jumps over the lazy dog"

(almost as famous as the "rain in spain falls only on the plain", though I don't know why this one is famous at all)

[Michael]

I'll just say that I'm correct here since Zarquon hasn't replied. I double checked and that sentence and it is 100% correct with every letter of the alphabet. :)

Here's anther puzzle...

There are three switches downstairs. Each corresponds to one of the three light bulbs in the attic. You can turn the switches on and off and leave them in any position. How would you identify which switch corresponds to which light bulb, if you are only allowed one trip upstairs?

[The Drunk Girl]

I'm sure this is not the answer you're looking for but... my lazy ass would wait for someone to get home and flick the switches for me downstairs while I was upstairs talking to them on my cell phone about which switch went to which light. :lol:

[drgoodtrips]

I'll just say that I'm correct here since Zarquon hasn't replied. I double checked and that sentence and it is 100% correct with every letter of the alphabet. :)

Here's anther puzzle...

There are three switches downstairs. Each corresponds to one of the three light bulbs in the attic. You can turn the switches on and off and leave them in any position. How would you identify which switch corresponds to which light bulb, if you are only allowed one trip upstairs?

I'd turn two of them on for a good long time. Then, I'd turn one of those off. When I went upstairs, the one that's on corresponds to the switch in the on position, the one that's hot and off corresponds to the switch newly in the off position, and the one that's cool and off corresponds to the switch that was always off.

[Zarquon]

I would leave two switches on, go upstairs, and see which bulb is off and if its only one, than the two switches lit up the two bulbs and the third one s the one linked with the off switch. if more than one is off, I would check the fuse and wiring while still upstairs, but wouldn't know how to proceed further.

[Michael]

I'd turn two of them on for a good long time. Then, I'd turn one of those off. When I went upstairs, the one that's on corresponds to the switch in the on position, the one that's hot and off corresponds to the switch newly in the off position, and the one that's cool and off corresponds to the switch that was always off.

Correct!

[drgoodtrips]

How about this:

One of these things is not like the others:

Boundaries
Cancerous
Librarian
Keyboards
Scorpions
Chameleon


Which word does not belong?

[Zarquon]

Cancerous as its an adjective, while the rest are nouns.

[Greendruid]

Hey look! He can tell the difference! :p Just kidding! I couldn't resist. Your honour is redeemed. Mine remains questionable :rofl:

[drgoodtrips]

Cancerous as its an adjective, while the rest are nouns.

Not the answer I was looking for, but true, so you're up. :lol:

[Michael]

How about this:

One of these things is not like the others:

Boundaries
Cancerous
Librarian
Keyboards
Scorpions
Chameleon


Which word does not belong?
1. Librarian (doesn't have an "o" in it).

2. The correct phrase should be "which one of this is not like the others"! :lol:

(yes, I know Zarquon is already correct, I'm just having fun!)

[drgoodtrips]

1. Librarian (doesn't have an "o" in it).

2. The correct phrase should be "which one of this is not like the others"! :lol:

(yes, I know Zarquon is already correct, I'm just having fun!)

Also not what I had in mind. The more esoteric answer is that all but "keyboards" contain symbols of the zodiac.

boundARIES
CANCERous
LIBRArian
SCORPIOns
chameLEOn

[Michael]

Also not what I had in mind. The more esoteric answer is that all but "keyboards" contain symbols of the zodiac.

boundARIES
CANCERous
LIBRArian
SCORPIOns
chameLEOn
Very cool! :cool:

I wonder if any other words contain zodiac names... :ummm:

(and my reference to the proper phrase should be accompanied by the Sesame Street tune in the background!)

[Zarquon]

If you encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

[dilettante]

If you encounter someone with exactly two children, given that at least one of them is a boy, what is the probability that both of her children are boys?

1 in 3.

The possible combinations are (GG), (BG), (GB), and (BB). Knowing that at least one is a boy rules out the (GG) possibility, leaving three.

[Michael]

1 in 3.

The possible combinations are (GG), (BG), (GB), and (BB). Knowing that at least one is a boy rules out the (GG) possibility, leaving three.

Since birth order is irrelevant to the question, there are only three possibilities and one of those possibilities is eliminated. Thus, one in two (50%).

[dilettante]

Since birth order is irrelevant to the question, there are only three possibilities and one of those possibilities is eliminated. Thus, one in two (50%).

That was my first thought too. But I think it's misleading (indeed, I bet this is what makes this a riddle instead of a really easy question).

Statistically (assuming that every child has a 50/50 shot at being a boy or girl), pairs of children are twice as likely to consist of a boy and a girl than to consist of two boys.
A fuller break down would be:

Child A is a boy (50% odds):
Child B is also a boy (50% of 50% = 25% odds)
Child B is a girl (50% of 50% = 25% odds)

Child A is a girl (50% odds):
Child B is a boy (50% of 50% = 25% odds)
Child B is also girl (50% of 50% = 25% odds) ruled out

[Zarquon]

Dilettante is right.

[dilettante]

A lost and hungry vagabond happened upon a pair of travelers one of whom had three loaves of bread while the other had five. All of the loaves were the same size and weight.

The two travelers decided to share their bread with the vagabond, and that the eight loaves should be shared equally among the three of them. When they had finished, the vagabond reached into his pocket and pulled out eight coins, all of equal value. He handed three coins to the traveler who had had the three loaves and five to the other one and disappeared into the inky shadows.

The next morning, right after no breakfast, the one who had received the three coins said to the other one, 'I don't think he should have given three coins to me and five to you. It's not fair.' And he was right. How should the coins have been split up?

[drgoodtrips]

7 coins to the high roller and 1 coin to his buddy.

[dilettante]

7 coins to the high roller and 1 coin to his buddy.

Wow. That one didn't last long.

Correct!
Each man ate 2 2/3 loaves, so the first man contributed 1/3 a loaf and the second 2 1/3 loaves to the vagabond. That's a ratio of 1-to-7.

[drgoodtrips]

For my brain teaser, I offer anyone the opportunity to help me debug my thread library in Operating Systems Design class to figure out why it seg faults on software driven interrupts.

Barring that, and since it's due at midnight, I should get back to work, so I'll defer my turn to anyone who wants to do a brain teaser... :erm:

[Michael]

For my brain teaser, I offer anyone the opportunity to help me debug my thread library in Operating Systems Design class to figure out why it seg faults on software driven interrupts.
Sounds like an elaborate form of "my dog ate my homework!" :lol:

[drgoodtrips]

Sounds like an elaborate form of "my dog ate my homework!" :lol:

That's my response here and to my teacher ;)

(You wouldn't believe what a tough sell that is with electronically submitted homework)

[Michael]

That's my response here and to my teacher ;)

(You wouldn't believe what a tough sell that is with electronically submitted homework)
A bug got my homework? My computer crashed? Virus infection? Malware?

I'm sure these excuses are giving the dog a run for the money! :lol:

[Michael]

Here's the next challenge:

I just found a number with an interesting property:

When I divide it by 2, the remainder is 1.
When I divide it by 3, the remainder is 2.
When I divide it by 4, the remainder is 3.
When I divide it by 5, the remainder is 4.
When I divide it by 6, the remainder is 5.
When I divide it by 7, the remainder is 6.
When I divide it by 8, the remainder is 7.
When I divide it by 9, the remainder is 8.
When I divide it by 10, the remainder is 9.

It's not a small number, but it's not really big, either.

Find the smallest number with such property.

[Zarquon]

2519.

[drgoodtrips]

A bug got my homework? My computer crashed? Virus infection? Malware?

I'm sure these excuses are giving the dog a run for the money! :lol:

Didn't need any. :D My group pulled off a coup, submitting our project literally at the 11th hour, and receiving a big, fat 100% from the auto-grading program. :bounce:

[Michael]

2519.

That's correct.

(It just occurred to me that this was a repeat question!) :o

[Zarquon]

~Dilettante and Greendruid are tied!~

Three Palefaces were taken captive by a hostile Indian tribe. According to tribe’s custom they had to pass an intelligence test, or die. The chieftain showed 5 headbands – 2 red and 3 white. The 3 men were blindfolded and positioned one after another, face to back. The chief put a headband on each of their heads, hid two remaining headbands, and removed their blindfolds. So the third man could see the headbands on the two men in front of him, the second man could see the headband on the first, and the first could not see any headbands at all.
According to the rules any one of the three men could speak first and try to guess his headband color. And if he guessed correctly – they passed the test and could go free, if not – they failed. It so happened that all 3 Palefaces were prominent logicians from a nearby academy. So after a few moments of silence, the first man in the line said: "My headband is ...".
What color was his head band? Why?

[drgoodtrips]

White.

The guy who could see two headbands remained silent, meaning that he couldn't tell which color his was by looking at the guys in front of him. This rules out the possibility that he was looking at two red ones, or he would have known his own was white.

So, the possibilities of the first two are reduced from:

RR
WR
RW
WW

to:

WR
RW
WW

Now, the second guy is aware of why the first guy isn't speaking, which means he knows that RR is not possible. He can see the headband in front of him, but he doesn't know which color his headband is. If he could see a red headband in front of him, he would know that the only possible way for that to happen would be WR, which means that his own headband is white. But, since he doesn't speak up, the first guy knows that the first and second headbands are not WR. So our remaining possibilities are:

RW
WW

In either case, the first guy's is W.

[Zarquon]

correct.

[drgoodtrips]

A man's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and the man died four years after his son.

How long did the man live?

[Michael]

A man's youth lasted one sixth of his life. He grew a beard after one twelfth more. After one seventh more of his life, he married. 5 years later, he and his wife had a son. The son lived exactly one half as long as his father, and the man died four years after his son.

How long did the man live?

84 years! :)

Btw, that "headband" puzzle sounds just like the "red hat vs white hat" puzzle.

[drgoodtrips]

84 years! :)

Btw, that "headband" puzzle sounds just like the "red hat vs white hat" puzzle.

Correct.

[Michael]

Okay, here's an easy one! :D

1. 26 L of the A
2. 7 D of the W
3. 7 W of the W
4. 12 S of the Z
5. 66 B of the B
6. 52 C in a P (W J)
7. 13 S in the U S F
8. 18 H on a G C
9. 39 B of the O T
10. 5 T on a F
11. 90 D in a R A
12. 3 B M (S H T R)
13. 32 is the T in D F at which W F
14. 15 P in a R T
15. 3 W on a T
16. 100 C in a D
17. 11 P in a F (S) T
18. 12 M in a Y
19. 13 is U F S
20. 8 T on an O
21. 29 D in F in a L Y
22. 27 B in the N T
23. 365 D in a Y
24. 13 L in a B D
25. 52 W in a Y
26. 9 L of a C
27. 60 M in an H
28. 23 P of C in the H B
29. 64 S on a C B
30. 9 P in the S S
31. 6 B to an O in C
32. 1000 Y in a M
33. 15 M on a D M C

Got to get them ALL to get the point! :D

Here's a hint on how it works: #1 is "26 Letters of the Alphabet"

[dilettante]

Well, it's been a week so I thought maybe we'd just make this one a group effort. I filled in all the ones that came off the top of my head in one pass. Anyone else want to fill in (some of) the rest?

1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 S of the Z
5. 66 Books of the Bible
6. 52 C in a P (W J)
7. 13 Stripes in the United States Flag
8. 18 H on a G C
9. 39 Books of the Old Testament
10. 5 T on a F
11. 90 D in a R A
12. 3 B M (S H T R)
13. 32 is the T in D F at which W F
14. 15 P in a R T
15. 3 W on a T
16. 100 C in a D
17. 11 P in a F (S) T
18. 12 Months in a Year
19. 13 is U F S
20. 8 T on an O
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 L of a C
27. 60 Minutes in an Hour
28. 23 P of C in the H B
29. 64 S on a C B
30. 9 P in the S S
31. 6 B to an O in C
32. 1000 Years in a Millennium
33. 15 M on a D M C

[drgoodtrips]

Well, it's been a week so I thought maybe we'd just make this one a group effort. I filled in all the ones that came off the top of my head in one pass. Anyone else want to fill in (some of) the rest?
1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 D in a R A
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Farenheit at which Water Freezes
14. 15 P in a R T
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 P in a F (S) T
18. 12 Months in a Year
19. 13 is U F S
20. 8 T on an O
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 L of a C
27. 60 Minutes in an Hour
28. 23 P of C in the H B
29. 64 S on a C B
30. 9 P in the S S
31. 6 B to an O in C
32. 1000 Years in a Millennium
33. 15 M on a D M C

[drgoodtrips]

I stopped at "cents in a dollar". I'll see if I can figure more out in a bit.

[dilettante]

1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 D in a R A
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Farenheit at which Water Freezes
14. 15 P in a R T
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 P in a F (S) T
18. 12 Months in a Year
19. 13 is U F S
20. 8 T on an O
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 Lives of a Cat
27. 60 Minutes in an Hour
28. 23 P of C in the H B
29. 64 S on a C B
30. 9 P in the S S
31. 6 B to an O in C
32. 1000 Years in a Millennium
33. 15 M on a D M C

[drgoodtrips]

1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 D in a R A
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Farenheit at which Water Freezes
14. 15 P in a R T
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 P in a F (S) T
18. 12 Months in a Year
19. 13 is Unlucky For Some???
20. 8 T on an O
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 Lives of a Cat
27. 60 Minutes in an Hour
28. 23 P of C in the H B
29. 64 S on a C B
30. 9 P in the S S
31. 6 B to an O in C
32. 1000 Years in a Millennium
33. 15 Men on a Dead Man's Chest

[The Drunk Girl]

I've been thinking about this shit since Michael posted it, and figured out roughly half of them. Thanks, Dilettante for suggesting this one be a group effort :) (Btw, Michael: D Y C U with T 33 on Y O?)


1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 Degrees in a Right Angle
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Farenheit at which Water Freezes
14. 15 P in a R T
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 P in a F (S) T
18. 12 Months in a Year
19. 13 is Unlucky For Some???
20. 8 Tentacles on an Octopus?!?!?:lol::shrug:
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 Lives of a Cat
27. 60 Minutes in an Hour
28. 23 P of C in the H B
29. 64 S on a C B
30. 9 P in the S S
31. 6 B to an O in C
32. 1000 Years in a Millennium
33. 15 Men on a Dead Man's Chest[/quote]

[Greendruid]

Yep, this is an oldie but a goodie - I remember seeing several versions of this since about age 8. I've provided two of the more obscure ones. We've used this as a document of culturally-specific knowledge in our intro classes before so I know I have the answers around here somewhere but this is more fun!

1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 Degrees in a Right Angle
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Farenheit at which Water Freezes
14. 15 P in a R T
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 P in a F (S) T
18. 12 Months in a Year
19. 13 is Unlucky For Some???
20. 8 Tentacles on an Octopus?!?!?:lol::shrug:
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 Lives of a Cat
27. 60 Minutes in an Hour
28. 23 Pairs of Chromosomes in the Human Body
29. 64 Squares on a Chess/Checker Board
30. 9 P in the S S
31. 6 Balls to an Over in Cricket
32. 1000 Years in a Millennium
33.15 Men on a Dead Man's Chest

[Zarquon]

1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 Degrees in a Right Angle
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Fahrenheit at which Water Freezes
14. 15 Players in a Rugby Team
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 Players in a Football (Soccer) Team
18. 12 Months in a Year
19. 13 is Unlucky For Some
20. 8 Tentacles on an Octopus
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 Lives of a Cat
27. 60 Minutes in an Hour
28. 23 Pairs of Chromosomes in the Human Body
29. 64 Squares on a Chess/Checker Board
30. 9 Planets in the Solar System
31. 6 Balls to an Over in Cricket
32. 1000 Years in a Millennium
33.15 Men on a Dead Man's Chest

[dilettante]

1. 26 Letters of the Alphabet
2. 7 Days of the Week
3. 7 Wonders of the World
4. 12 Signs of the Zodiac
5. 66 Books of the Bible
6. 52 Cards in a Pack (Without Jokers)
7. 13 Stripes in the United States Flag
8. 18 Holes on a Golf Course
9. 39 Books of the Old Testament
10. 5 Toes on a Foot
11. 90 Degrees in a Right Angle
12. 3 Blind Mice (See How They Run)
13. 32 is the Temperature in Degrees Fahrenheit at which Water Freezes
14. 15 Players in a Rugby Team
15. 3 Wheels on a Tricycle
16. 100 Cents in a Dollar
17. 11 Players in a Football (Soccer) Team
18. 12 Months in a Year
19. 13 is Unlucky For Some
20. 8 Tentacles on an Octopus
21. 29 Days in February in a Leap Year
22. 27 Books in the New Testament
23. 365 Days in a Year
24. 13 Loaves in a Baker's Dozen
25. 52 Weeks in a Year
26. 9 Lives of a Cat
27. 60 Minutes in an Hour
28. 23 Pairs of Chromosomes in the Human Body
29. 64 Squares on a Chess/Checker Board
30. 9 Planets in the Solar System
31. 6 Balls to an Over in Cricket
32. 1000 Years in a Millennium
33.15 Men on a Dead Man's Chest

Nice! I think that's all of them. Good work folks. :)

[Michael]

Nice! I think that's all of them. Good work folks. :)

Indeed - all of them. Only one "error" on the list. Shall I give the point to the one who correctly guesses the error? :rofl:

[drgoodtrips]

Is it the one I put a question mark next to after answering, "13 is unlucky for some"?

[Michael]

Is it the one I put a question mark next to after answering, "13 is unlucky for some"?

Nope. That one is correct. :D

[The Drunk Girl]

Is it that there really isn't considered 9 planets in the solar system anymore? :shrug:

[drgoodtrips]

Ooh, good one!

[Michael]

Is it that there really isn't considered 9 planets in the solar system anymore? :shrug:
The actual number of scientifically defined planets may be questionable, but the "9 planets in the solar system" is certainly the correct solution to the question of "9 P in the SS". ;)

I'll give a hint. It is one of the "sports" related questions and involves a sport that most people here likely know little about. :)

[Zarquon]

11 Players in a Football (Standard) Team?

[Michael]

11 Players in a Football (Standard) Team?

No, that one is correct with (Soccer). :D

[The Drunk Girl]

I'll give a hint. It is one of the "sports" related questions and involves a sport that most people here likely know little about. :)

I will assume that it has something to do with croquet instead of cricket? Beyond that I have no clue to how either are played and what is involved other than their names.

[Michael]

I will assume that it has something to do with croquet instead of cricket? Beyond that I have no clue to how either are played and what is involved other than their names.
Well, that should make it easy then since there are no clues about croquet. ;)

[Zarquon]

there are 6 balls in an over in cricket.
besides that i don't know what you're getting at.

[Michael]

there are 6 balls in an over in cricket.
besides that i don't know what you're getting at.

You got the right one with the wrong error! :lol:

It should be "6 bowls to an over in cricket".

I'll give 1/2 point to all the contributors to the solution on this one... dilettante, drgoodtrips, Drunk Girl, Greendruid and Zarquon! :)

[Zarquon]

You got the right one with the wrong error! :lol:

It should be "6 bowls to an over in cricket".

I'll give 1/2 point to all the contributors to the solution on this one... dilettante, drgoodtrips, Drunk Girl, Greendruid and Zarquon! :)
http://en.wikipedia.org/wiki/Cricket_ball
It is ball in cricket; bowl is not used, except in the term of being bowled out (http://en.wikipedia.org/wiki/Bowled).

[Michael]

http://en.wikipedia.org/wiki/Cricket_ball
It is ball in cricket; bowl is not used, except in the term of being bowled out (http://en.wikipedia.org/wiki/Bowled).
That is true.

But there certainly are six "bowls" to an over, not six "balls" to an over.

And I didn't make up the quiz. ;)

[Zarquon]

That is true.

But there certainly are six "bowls" to an over, not six "balls" to an over.

And I didn't make up the quiz. ;)
A 'bowl' is a utensil, and thus not present in an over. It really is six balls to an over. Look it up anywhere, or just refer to the link I've posted.

Anyway, who's supposed to post the next riddle?

[Michael]

A 'bowl' is a utensil, and thus not present in an over. It really is six balls to an over. Look it up anywhere, or just refer to the link I've posted.
Bowling is the word for 'pitching' in cricket. Yes it is a ball, but it is "bowled" down the pitch to the batter (at extremely high speed!).

Anyway, who's supposed to post the next riddle?
Same as always - whoever is ready and willing. Please feel free.

[Michael]

Here's a puzzle...

How many mice are there in the room if there is a mouse in each of the four corners, and opposite to each mouse there are three mice and at each mouse's tail, there is a mouse?

[Zarquon]

four only

[Michael]

four only

You are correct!

[Greendruid]

This falls more into the realm of trivia I suppose but I'll Michael adjudicate on allowing here.

You can probably name one, maybe even two but can you name three of the non-Indo-European languages that are still spoken today in the European continent? We will use as the definition of the European continent everything west of the Russian border that is contiguous (i.e., excluding Asia Minor). Remember the rules kids - no checking the inter-webs!

[Michael]

This falls more into the realm of trivia I suppose but I'll Michael adjudicate on allowing here.
Sure - whynot? Consider the title to be "Brain Teasers, Riddles, Puzzles and Trivia" Challenge.

You can probably name one, maybe even two but can you name three of the non-Indo-European languages that are still spoken today in the European continent? We will use as the definition of the European continent everything west of the Russian border that is contiguous (i.e., excluding Asia Minor). Remember the rules kids - no checking the inter-webs!
I'll guess Finnish, Hungarian and Basque.

[Greendruid]

Sure - whynot? Consider the title to be "Brain Teasers, Riddles, Puzzles and Trivia" Challenge.


I'll guess Finnish, Hungarian and Basque.

Michael is correct!

[Michael]

Okay... been a while here...

I got this one from my local radio station - they have a feature called the "impossible question of the day".

Apparently the common flu virus can live on this common item/object for up to 17 days. What is that common item/object?

[drgoodtrips]

A water fountain?

[Michael]

A water fountain?

Nope!

[The Drunk Girl]

Sink handles/toilet handles?

[The Drunk Girl]

Scratch that

[Michael]

Sink handles/toilet handles?
Nope.

I should think that smooth metal objects wouldn't be very good 'hosts' for such microbe/germs/bacteria/virii. Certainly possible, but not ideal hosts. That's just my inclination of course - I'm woefully weak in biology topics. :o

[Michael]

Scratch that
No harm in guessing. You can try again. This isn't Jeopardy so you don't lose $400 for giving an incorrect answer (or question!). :D

And yes, I'm a huge fan of Jeopardy. I grew up watching that show because my mother loved it. Its been on the air as long as I've been alive it seems. And I've been told that I really ought to try out for that show. If I like the category, I can usually get all five questions in that category - I usually get about 1/3 of all the questions on that show. :shrug:

[drgoodtrips]

I watch a lot of jeopardy too. That's fun show.

[Margot]

Okay... been a while here...

I got this one from my local radio station - they have a feature called the "impossible question of the day".

Apparently the common flu virus can live on this common item/object for up to 17 days. What is that common item/object?

I think I read this in a Cracked article. Is it a keyboard?

[The Drunk Girl]

Nope.

I should think that smooth metal objects wouldn't be very good 'hosts' for such microbe/germs/bacteria/virii. Certainly possible, but not ideal hosts. That's just my inclination of course - I'm woefully weak in biology topics. :o

Originally I was going to say fomites, but when I saw what dr. said I knew mine wouldn't work. Hence me saying, "scratch that" ;)

[The Drunk Girl]

No harm in guessing. You can try again. This isn't Jeopardy so you don't lose $400 for giving an incorrect answer (or question!). :D

And yes, I'm a huge fan of Jeopardy. I grew up watching that show because my mother loved it. Its been on the air as long as I've been alive it seems. And I've been told that I really ought to try out for that show. If I like the category, I can usually get all five questions in that category - I usually get about 1/3 of all the questions on that show. :shrug:

If you get on you know you have to let us know. That just wouldn't be fair :p

[Michael]

I think I read this in a Cracked article. Is it a keyboard?
Nope!

If you get on you know you have to let us know. That just wouldn't be fair :p
Maybe.

Btw, I was in the official queue to be a contestant for "Who Wants to be a Millionaire" when that show was cancelled. :shrug:

[Michael]

Oddly enough, I thought this question was really easy when I heard it on the radio. I only posted it because the radio call-in's had a real hard time with it. I got it correct with my first guess!

[Margot]

Okay... been a while here...

I got this one from my local radio station - they have a feature called the "impossible question of the day".

Apparently the common flu virus can live on this common item/object for up to 17 days. What is that common item/object?

Pillow, nostril, lamp switch, steering wheel, windscreen, fingernail, UNDER fingernail, pool, vagina, elephant, microwave, spoon washed on "light wash" setting to be eco-friendly, spouse, small child, aglet, in the vent of a heater in south Florida directly over a humidifier in the room of a small child who often picks his nose and wipes his findings on the aglets of his laces?

Am I getting anywhere close? RUMPELSTILTSKIN!

[The Drunk Girl]

Pillow, nostril, lamp switch, steering wheel, windscreen, fingernail, UNDER fingernail, pool, vagina, elephant, microwave, spoon washed on "light wash" setting to be eco-friendly, spouse, small child, aglet, in the vent of a heater in south Florida directly over a humidifier in the room of a small child who often picks his nose and wipes his findings on the aglets of his laces?

Am I getting anywhere close? RUMPELSTILTSKIN!

:rofl:

Telephone

[Michael]

Pillow, nostril, lamp switch, steering wheel, windscreen, fingernail, UNDER fingernail, pool, vagina, elephant, microwave, spoon washed on "light wash" setting to be eco-friendly, spouse, small child, aglet, in the vent of a heater in south Florida directly over a humidifier in the room of a small child who often picks his nose and wipes his findings on the aglets of his laces?

Am I getting anywhere close? RUMPELSTILTSKIN!
None of the above! :D

Not even close actually. Spouse or small child is a good guess though!

And I'd forgotten what the plastic thing at the end of one's laces was called! Aglets!!! :rofl:

[SMadsen]

The next question ought to be who invented aglets :)

Anyhow, everyone knows that paper money are regular flu zoo's ;)

[Margot]

The next question ought to be who invented aglets :)

Anyhow, everyone knows that paper money are regular flu zoo's ;)

I forgot all about money! I suspect it's because I never have any (and when I do, it always goes through the wash at least once via back pocket).

[Michael]

The next question ought to be who invented aglets :)

Anyhow, everyone knows that paper money are regular flu zoo's ;)
SMadsen is correct. :)

[drgoodtrips]

SMadsen is correct. :)

Poor cokeheads...

[Michael]

Next question is....

Who invented aglets? :lol:

[Michael]

Haven't posted one of these in a while...

Q: One of the people from Mr. A to Mr. F is lying. Who is it?

A: I know one of us is lying.
B: I am not a liar.
C: Either D or B is honest.
D: I know A or C is telling the truth.
E: B is honest.
F: I am sure there is a liar.

[WFCY]

Haven't posted one of these in a while...

Q: One of the people from Mr. A to Mr. F is lying. Who is it?

A: I know one of us is lying.
B: I am not a liar.
C: Either D or B is honest.
D: I know A or C is telling the truth.
E: B is honest.
F: I am sure there is a liar.

Well, assuming "or" used above is consistently the exclusive or (http://en.wikipedia.org/wiki/Exclusive_or).

D is the liar.

Reasoning goes like this- Assume B is honest. This makes A, E, and F consistent. For C, D must be the one who is lying if C is telling the truth. This is a possibility by looking at what D said: either if A and C are both lying, or if A and C are both telling the truth, either one is sufficient condition to qualify D as a liar. Since A and C are both truthful, D is the liar.

If you think in terms of the inclusive or, then the result may seem very unintuitive. But suppose if you ask me: "who is the liar", and suppose I know both A and B are, and said "either A or B is a liar", I would have been lying there. That's another instance of exclusive or usage.

[Michael]

I think if you use exclusive OR, then C or D are equally likely.

If you use inclusive OR (i.e. and/or), then one answer becomes clear.

(btw, there was a long and extensive debate about this riddle at the place where I found it)

[WFCY]

I think if you use exclusive OR, then C or D are equally likely.

If you use inclusive OR (i.e. and/or), then one answer becomes clear.

(btw, there was a long and extensive debate about this riddle at the place where I found it)

if we use inclusive or, then A is the liar.

[Michael]

That sure took a while for me to find... :o

Anyway, here's the answer that came with the riddle and it makes sense to me. The poster of the riddle stated that this was 'the best' answer he'd seen to the riddle, but didn't claim it to be the definitive answer.

Mr. C

We know there is a liar thanks to the question itself. That means Mr. A & Mr. F are telling the truth.

Mr. E and Mr. B both agree that Mr. B is not a liar. Since there is only one liar, they are both truthful.

That leaves Mr. D and Mr. C

Mr. D says that he knows Mr. A OR Mr. C is telling the truth. Since we know Mr. A is indeed telling the truth, that makes Mr. D also a truth teller, and points to Mr. C being the liar, as both Mr. B and Mr. D have already both been found to be honest.

I agree that your answer also works, so I'll certainly give you that. Your name will be duly listed on the scoring list (see page1). :)

Since you have a correct answer, the next challenge is yours to post! :D

[drgoodtrips]

I personally think that is paradoxical. Assuming a truth equivalence of "is" "I know", "I am sure", requires the number of untrue statements not to be equal to 1. Logically, F equiv A and E equiv B, meaning those are package deals (either zero or 2 false statements). D and C both give logical or, which can only be false with 2 or more false statements.

In the 'solution' C, the "liar", has a statement that is equivalent to: "B or D". If C is, in fact, a liar (making a false statement), the constraints of the original problem are violated as this makes both B and D liars (as well as E, then A, then F, respectively).

In other words, the answer to the riddle is that the narrator is the liar. ;)

[Michael]

I personally think that is paradoxical. Assuming a truth equivalence of "is" "I know", "I am sure", requires the number of untrue statements not to be equal to 1. Logically, F equiv A and E equiv B, meaning those are package deals (either zero or 2 false statements). D and C both give logical or, which can only be false with 2 or more false statements.

In the 'solution' C, the "liar", has a statement that is equivalent to: "B or D". If C is, in fact, a liar (making a false statement), the constraints of the original problem are violated as this makes both B and D liars (as well as E, then A, then F, respectively).

In other words, the answer to the riddle is that the narrator is the liar. ;)

Yes, the site where I got that from, the majority consensus was "C", but there was a significant minority asserting that the 'narrator' was guilty.

[WFCY]

C and D are equally possible if we use the exclusive or, A is the only possible one (besides the narrator if that's even allowed), becuase maybe he does not "know", he just says he does, if we use inclusive or. Otherwise I cant see a solution.

I have not read thru all the pages so maybe I am asking a question twice.


You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.
What is the smallest number of times you must use the scale in order to always find the fake coin?
Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.
These are modern coins, so the fake coin is not necessarily lighter.
Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt.

[Michael]

You have twelve coins...

Stumps me entirely. I had to google it, got the answer and I still don't get it! :lol:

This one is very challenging to say the least! :erm:

Where's drgoodtrips? (or Dominick) They are the only ones whom I would guess would be likely to figure this one out. :shrug:

[WFCY]

Stumps me entirely. I had to google it, got the answer and I still don't get it! :lol:

This one is very challenging to say the least! :erm:

Where's drgoodtrips? (or Dominick) They are the only ones whom I would guess would be likely to figure this one out. :shrug:

yeah, it's a hard one, but the answer is really quite fun and challenging once you understand it.

I figured it out after 2 hints (or 1, its about having have to mix balls) and several days. I loved it.

[dilettante]

You have twelve coins. You know that one is fake. The only thing that distinguishes the fake coin from the real coins is that its weight is imperceptibly different. You have a perfectly balanced scale. The scale only tells you which side weighs more than the other side.
What is the smallest number of times you must use the scale in order to always find the fake coin?
Use only the twelve coins themselves and no others, no other weights, no cutting coins, no pencil marks on the scale. etc.
These are modern coins, so the fake coin is not necessarily lighter.
Presume the worst case scenario, and don't hope that you will pick the right coin on the first attempt.

I can manage it in 4 weighings a couple of ways...still trying to think of a way to do it in three.


Four weighing method:
1) Weigh three against three with six off the scale. If it tilts, the coin is among the six being weighed; if not it is among the six off the scale. This drops the possibilities down to six.
2) Weigh three of the possible six against three confirmed good coins. If it tilts you know the bad coin is on the scale (and you know whether its lighter or heavier); if it doesn't tilt, the bad coin is off the scale. Either way, you're down to three.
3) Of the three remaining weigh two against each other. If the scale tilts and you know (from step 2) whether the bad coin is heavier or lighter, you get the answer right here. If it doesn't tilt, the bad coin is the one yet to be weighed. if it tilts and you don't know whether the bad one is heavier or lighter, you have two remaining possibilities.
4) Weigh one of the remaining two against a good coin. If it tilts, its the bad coin. if not, the other possibility is bad.

[Michael]

I can manage it in 4 weighings a couple of ways...still trying to think of a way to do it in three.
My apologies for not including you on the list of likely solvers. :o

Your leading score here is obviously for a reason. :)

[dilettante]

Well, I'm just going to assume that I was right about that last one and post something new. :) Here's something different. Technically, this is a paradox, not a riddle, but it still might be fun to chew on mentally. A point for the person who demonstrates its paradoxical nature?


The Two Envelope Paradox

A crazy rich uncle places two identical envelopes before you and tells you that each envelope contains a check made out to you. One check is for twice as much money as the other, but he doesn't tell you which is which or what the amounts are. You can take one of the envelopes and the check inside, but the other will be destroyed.

Choosing one envelope at random, you open it and find that the check inside is for $500. At this point, your crazy uncle gives you the option of tearing up that check and taking the other envelope instead. Are you more likely to gain or lose money by taking the other envelope? Or does it matter?

[Michael]

Well, I'm just going to assume that I was right about that last one and post something new. :)
Well, that would be WFCY's call since he posted the challenge. :)

As I mentioned above, I gave up and searched it on the net - I believe that a 'better' answer does exist, but as I noted then and WFCY admitted, this is a very challenging puzzle. :ummm:

Here's something different. Technically, this is a paradox, not a riddle, but it still might be fun to chew on mentally. A point for the person who demonstrates its paradoxical nature?


The Two Envelope Paradox

A crazy rich uncle places two identical envelopes before you and tells you that each envelope contains a check made out to you. One check is for twice as much money as the other, but he doesn't tell you which is which or what the amounts are. You can take one of the envelopes and the check inside, but the other will be destroyed.

Choosing one envelope at random, you open it and find that the check inside is for $500. At this point, your crazy uncle gives you the option of tearing up that check and taking the other envelope instead. Are you more likely to gain or lose money by taking the other envelope? Or does it matter?
I'd say that it is a paradox because one 'gains' no matter what, regardless of the result, since any gifted check is better than nothing that one started with - even if you 'lose' the opportunity for a relative or potentially greater 'gain'.

[amberson]

Ohh..my brain is really twisting on the riddles. I cant find the answer.

[dilettante]

The Two Envelope Paradox

A crazy rich uncle places two identical envelopes before you and tells you that each envelope contains a check made out to you. One check is for twice as much money as the other, but he doesn't tell you which is which or what the amounts are. You can take one of the envelopes and the check inside, but the other will be destroyed.

Choosing one envelope at random, you open it and find that the check inside is for $500. At this point, your crazy uncle gives you the option of tearing up that check and taking the other envelope instead. Are you more likely to gain or lose money by taking the other envelope? Or does it matter?

Ha. I forgot all about this thread. For the sake of achieving closure:

The paradox arises when one attempts to mathematically calculate the expected reward from switching envelops:

Your first envelop had a $500 check inside. Ergo, there is a 50% chance that switching envelops will give you a $1000 check instead and a 50% chance that switching will give you a $250 check. Thus, the expected reward for switching is:
(1/2)($1000) + (1/2)($250) = $500 + $125 = $625
Which is more than the $500 you get for keeping your first choice. Ergo, statistically, you should switch: the reward outweighs the risk.

However, the statistics work out the same way no matter what you find in the first envelope. Whatever the amount, you have a 50% chance of increasing it by 100% and a 50% percent chance of decreasing it by 50%, so the average return from switching will always be a gain of 25% over whatever you started with:
(1/2)(2*X) + (1/2)(X/2) = X+ X/4 = 1.25*X
So you really don't even need to open the first envelope for it to make sense, statistically, to switch to the second.

But, since you chose randomly in the first place, the odds of picking the larger check are the same as the odds of picking the smaller check. So switching checks can't possibly improve your situation at all; it's just a 50/50 shot either way.

Both evaluations are, statistically, correct; yet they result in conflicting results. Hence the paradox.

And I'd still like to know what the better answer to WFCY's puzzle was.

[drgoodtrips]

They're not both statistically correct. The first analysis is shifting the goalposts subtly. I'm off to a meeting, but I'll offer an explanation later -- it has to do with explanation (1) not taking into account the contents of both envelopes.

[drgoodtrips]

Alright, so by way of explanation. Consider that with the first scenario (expected value of $625), assuming that it's correct, you could use it on this 'same' problem:

"Take all of the money out of your wallet. Would you like to play a game where you flip a coin and, if it's heads, I double your money, and if it's tails, I take half of your money."

In this game, your expected value is 1.25 times your money, so if you're rational (and not risk averse) you play.

Problem is, these two games are not the same. It doesn't pass the common sense test - there's no mention of switching and there aren't two mysterious quantities.

In the game with the two envelopes, you pick one and then are asked if you want to switch. But the phrasing of the expected value proposition ignores the fact that you have already picked an envelope.

Your envelopes have value x and 2x. When you make your first pick, there is a 50 percent chance that you've picked x and a 50% chance that you've picked 2x, so your intermediate expected value is 1.5x. You're starting with 1.5x rather than (x or 2x). So, when you go to switch, there's a 50% chance that you'll lose .5x and a 50% chance that you'll gain .5x because the switch means that you've either gone from 1.5x to either 2x or x.

The problem isn't with the statistics, but with how the problem is described. In scenario one, you can't compare the expected values because of the shifting goalposts. If we have an expected value scenario where a heads tails coin flip results in different amounts of money, we can do 5.* heads amount + .5 * tails amount. This expected value model is applied here as well, but it is not appropriate because the parameters are not the same. With each outcome, we assume a different state of the world before the game is played. We say, there's a 50% chance that x1 happened and a 50% chance that x2 happened, so we'll calculate expected value with .5*y1 and .5*y2. That's not how expected value works - the goalposts are being shifted.

Perhaps an easier way to understand it is to say that the game of switching envelopes is a zero sum game. There is some fixed amount of money in the envelopes, represented by X and 2X. If two people are adversaries in this game, the amount of money is fixed, and one will wind up with more at the expense of the other. In the 'paradoxical' scenario, the riddle-teller sets up a zero sum game and then pulls the switcheroo by offering an open ended game and challenging the reader to reconcile the concepts. Switching envelopes is zero sum and the money is fixed. Giving someone money and then saying 50% chance of doubling and 50% chance of halving is not a zero sum game because it's unknown how much money there will be in the end.

So, the 'paradox' is like saying "I have two colored objects behind my back, a red one and a blue one. If you pick one and I show you that it's red, what are the odds that the next one you pick will be blue? Also, there are now three total objects."

[drgoodtrips]

It just occurred to me that there's an easier way to describe the situation. The answer to the riddle is that you're equally likely to gain as you are to lose, per the second explanation. The first explanation describes a different problem and is not applicable to the 'game' in question (of switching envelopes).

[dilettante]

It just occurred to me that there's an easier way to describe the situation. The answer to the riddle is that you're equally likely to gain as you are to lose, per the second explanation. The first explanation describes a different problem and is not applicable to the 'game' in question (of switching envelopes).

That works for me.
Here's a more regular brain teaser to compensate for that last entry:

Most digital clocks give the hour and the minute, but not the second. At various times of day, such as five minutes before six, such a clock will display three of the same digit. For example 5:55.
How many times during the course of a complete day (24-hours) does a digital clock display a time that has three of the same digit?

[The Drunk Guy]

That works for me.
Here's a more regular brain teaser to compensate for that last entry:

Most digital clocks give the hour and the minute, but not the second. At various times of day, such as five minutes before six, such a clock will display three of the same digit. For example 5:55.
How many times during the course of a complete day (24-hours) does a digital clock display a time that has three of the same digit?
36 times

[dilettante]

36 times

Nope. More.

[Greendruid]

I'm coming up with 48.

[dilettante]

I'm coming up with 48.

Bing! I believe that's correct. :)

12:11, 12:22
1:11
2:22
3:33
4:44
5:55
10:00, 10:11
11:01, 11:10 - 11:19, 11:21, 11:31, 11:41, 11:51

[Greendruid]

I made the exact same mistake as TDGuy first too! The 11s were tricky. Here is the next brain-teaser ...

I recently read one of those books that are so good that you just can't put it down. When I started reading I noticed that both hands on my wall clock were exactly on minute marks, and when I finally finished the book eight hours later, both hands on the clock were again exactly on minute marks, but the hour hand was now six times as far ahead of the minute hand as it had been when I started - going clock-wise.

What time did I start reading the book?

[The Drunk Guy]

I made the exact same mistake as TDGuy first too! The 11s were tricky.
Actually, my mistake was trying to tackle a problem before noon. :lol:

[Greendruid]

Actually, my mistake was trying to tackle a problem before noon. :lol:

I stand corrected then ... almost the same mistake :D

[dilettante]

I recently read one of those books that are so good that you just can't put it down. When I started reading I noticed that both hands on my wall clock were exactly on minute marks, and when I finally finished the book eight hours later, both hands on the clock were again exactly on minute marks, but the hour hand was now six times as far ahead of the minute hand as it had been when I started - going clock-wise.

What time did I start reading the book?

For some reason, I found this puzzle incredible difficulty to wrap my head around.

Anyway, after much annoyance, one possible answer is: 6:24.

6:24 ==> Minute hand on the 24, hour hand exactly on the 32 minute mark = 8 ticks apart
+8 hrs
2:24 ==> Minute hand on the 24, hour hand exactly on the 12 minute mark = 48 ticks apart (going clockwise)

48 = 8 * 6

[Greendruid]

For some reason, I found this puzzle incredible difficulty to wrap my head around.

Anyway, after much annoyance, one possible answer is: 6:24.

6:24 ==> Minute hand on the 24, hour hand exactly on the 32 minute mark = 8 ticks apart
+8 hrs
2:24 ==> Minute hand on the 24, hour hand exactly on the 12 minute mark = 48 ticks apart (going clockwise)

48 = 8 * 6

Ding! Ding! We have a winnuh! The honour of the next teaser falls to you, good sir.

[dilettante]

Here's the new one:

There are two telephone poles. Each one is 100-feet tall. They are parallel and an unknown distance apart.

We're going to attach a 150-foot rope from the very top of one of the poles to the top of the other. This rope will, of course, droop down somewhat. That drooping rope is called a catenary, from the Latin word for chain.

The question is: What must be the distance between the two poles, so that the lowest point of the catanary is 25-feet above the ground?

[drgoodtrips]

Won't that depend on the tension and density of the rope?

[dilettante]

Won't that depend on the tension and density of the rope?

For the purposes of this puzzle, assume an ideal "rope" that maintains a fixed length of 150 ft regardless of the stresses placed upon it.

[Dominick]

Here's the new one:

There are two telephone poles. Each one is 100-feet tall. They are parallel and an unknown distance apart.

We're going to attach a 150-foot rope from the very top of one of the poles to the top of the other. This rope will, of course, droop down somewhat. That drooping rope is called a catenary, from the Latin word for chain.

The question is: What must be the distance between the two poles, so that the lowest point of the catanary is 25-feet above the ground?
Is this a trick question (with me missing the trick) or do you really want people to mess with hyperbolic functions ? :D

[dilettante]

Is this a trick question (with me missing the trick) or do you really want people to mess with hyperbolic functions ? :D

It isn't a trick question (at least I don't consider it as such). And one could certainly use advanced math to solve it and come up with the correct answer.

But, as it happens, one can also solve it using extremely simple 'in-your-head' math.

Edit: I'd actually forgotten all about this, or I would have just given the answer away for the sake of moving the game along. I'll wait a day in case you want to take a stab at it. Then I might as well give the answer since it's 3+ months old. :o

[Dominick]

It isn't a trick question (at least I don't consider it as such). And one could certainly use advanced math to solve it and come up with the correct answer.

But, as it happens, one can also solve it using extremely simple 'in-your-head' math.

Edit: I'd actually forgotten all about this, or I would have just given the answer away for the sake of moving the game along. I'll wait a day in case you want to take a stab at it. Then I might as well give the answer since it's 3+ months old. :o
I spent several hours with integrals, trigonometric and hyperbolic functions while cursing you until it hit me. :rofl:

The answer is of course ...... zero.

[Greendruid]

I spent several hours with integrals, trigonometric and hyperbolic functions while cursing you until it hit me. :rofl:

The answer is of course ...... zero.

Hilarious! We all missed this one it seems, or at least forgot about it. The honour of posting the next puzzle surely goes to Dominick but I'll wait for Dil to confirm the answer

[drgoodtrips]

I spent several hours with integrals, trigonometric and hyperbolic functions while cursing you until it hit me. :rofl:

The answer is of course ...... zero.

:facepalm: :o

[kate]

Hmmm sounds tricky but again brainteasers are meant to be tricky enough to tease the brain!lol.I think it doesn't really matter which boy speaks up , lets say the boy who always lie points to a wrong direction and the man follows it and if he doesn't finds a way he would obviously follow the opposite direction which is supposed to be the correct route.
Oh I know this one isn't correct but I seriously can't figure out until you give me a hint.:shrug:
Check this teaser -
www.iqtestexperts.com/brainteasers/

[drgoodtrips]

This is kind of a mindfuck:

http://beust.com/weblog/2011/10/27/probability-quiz/

I'll have to think about this more when I'm not quite so busy.

[Dominick]

I'm going with 'Undefined'. It should be rephrased to make it a genuine probability question.

[NickKIELCEPoland]

It isn't a trick question (at least I don't consider it as such). And one could certainly use advanced math to solve it and come up with the correct answer.

But, as it happens, one can also solve it using extremely simple 'in-your-head' math.

Edit: I'd actually forgotten all about this, or I would have just given the answer away for the sake of moving the game along. I'll wait a day in case you want to take a stab at it. Then I might as well give the answer since it's 3+ months old. :o
This post was edited on the 5th of October - and no word since, from Dillitante.
Well, except on other threads, where dillitante has been very active.

[dilettante]

This post was edited on the 5th of October - and no word since, from Dillitante.
Well, except on other threads, where dillitante has been very active.

Dominick answered it in post #458. :shrug:
It was so easy to do the math and see he was right, and everyone seemed to acknowledge he was right; I didn't think it was necessary to state the obvious.

But yes, "bing!" 0 is the right answer. Sorry if the delay caused confusion.

[NickKIELCEPoland]

Well done, Dominick :)

[NickKIELCEPoland]

What -ism began and ended with the execution of a person who was named after the aim of that -ism?

[Michael]

Dominick answered it in post #458.

Did you score a credit to Dominick on the first page?

[drgoodtrips]

I'm going with 'Undefined'. It should be rephrased to make it a genuine probability question.

I'm with you, actually. There's nothing in the question to indicate that the percentages aren't just meaningless values or answers to another question.

[dilettante]

Did you score a credit to Dominick on the first page?

I have now. :)

[dilettante]

This thread has been quiet for a long time and I doubt Dominick is going to post the next riddle, so here's one to possibly get things going again:

A box contains three cards. One card is green on both sides. One card is red on both sides. The third card is green on one side and red on the other. The cards are placed in the box and the box is shaken. One card is randomly drawn out and placed on the table. The face-up side of the card is red.

Question: What are the odds that the other side of the card is also red?

[NickKIELCEPoland]

This thread has been quiet for a long time and I doubt Dominick is going to post the next riddle, so here's one to possibly get things going again:

A box contains three cards. One card is green on both sides. One card is red on both sides. The third card is green on one side and red on the other. The cards are placed in the box and the box is shaken. One card is randomly drawn out and placed on the table. The face-up side of the card is red.

Question: What are the odds that the other side of the card is also red?
2 / 3 is my suggestion.

[Greendruid]

This thread has been quiet for a long time and I doubt Dominick is going to post the next riddle, so here's one to possibly get things going again:

A box contains three cards. One card is green on both sides. One card is red on both sides. The third card is green on one side and red on the other. The cards are placed in the box and the box is shaken. One card is randomly drawn out and placed on the table. The face-up side of the card is red.

Question: What are the odds that the other side of the card is also red?

50/50. At the point that you have the red side up established, the green card can be ignored in the finite math calculation of the odds of whether it is the half green/half red card or the all-red card that was drawn out of the box.

[dilettante]

2 / 3 is my suggestion.

50/50. At the point that you have the red side up established, the green card can be ignored in the finite math calculation of the odds of whether it is the half green/half red card or the all-red card that was drawn out of the box.

Nick has it right, I believe.

True, there are two possible cards (the red-red and the red-green), but the card on the table is more likely to be the red-red.

It helps to think in terms of sides, rather than of cards. That is to say, when you see which side is face up on the table you can list the possibilities as such:
1) Side A of the green-green card
2) Side B of the green-green card
3) The green side of the red-green card
4) The red side of the red-green card
5) Side A of the red-red card
6) Side B of the red-red card

We can eliminate the first 3 possibilities because the side we can see is red, not green. That leaves three equally probable possibilities left and for 2 or the 3 the other side of the card is red.

Your turn to post a riddle/puzzle, Nick.

[NickKIELCEPoland]

Nick has it right, I believe.

True, there are two possible cards (the red-red and the red-green), but the card on the table is more likely to be the red-red.

It helps to think in terms of sides, rather than of cards. That is to say, when you see which side is face up on the table you can list the possibilities as such:
1) Side A of the green-green card
2) Side B of the green-green card
3) The green side of the red-green card
4) The red side of the red-green card
5) Side A of the red-red card
6) Side B of the red-red card

We can eliminate the first 3 possibilities because the side we can see is red, not green. That leaves three equally probable possibilities left and for 2 or the 3 the other side of the card is red.

Your turn to post a riddle/puzzle, Nick.
I will simply repeat the one I've already posted. (no one has tried to solve it)
Which ideology's heyday began and ended with the execution of someone whose name was synonymous with that ideology's stated aim?

[dilettante]

I will simply repeat the one I've already posted. (no one has tried to solve it)
Which ideology's heyday began and ended with the execution of someone whose name was synonymous with that ideology's stated aim?

I give up. :shrug:

[NickKIELCEPoland]

I give up. :shrug:
Well at least you tried - everyone else has just poo-poo'd it thus far.
I will give everyone a couple of days to see if they can get it - it really shouldn't be too difficult for politico-historical animals.

[dilettante]

I will simply repeat the one I've already posted. (no one has tried to solve it)
Which ideology's heyday began and ended with the execution of someone whose name was synonymous with that ideology's stated aim?

I still give up.

[Michael]

I will simply repeat the one I've already posted. (no one has tried to solve it)
Which ideology's heyday began and ended with the execution of someone whose name was synonymous with that ideology's stated aim?

That's a pretty short time period - if it begins and ends with a single event. That means it ended immediately the moment it began.

I'd say that's a time period too short to be considered relevant for anything.

[NickKIELCEPoland]

Michael, the time period was between two executions of men whose name was synonymous with that ideology's stated aim.

[Greendruid]

Michael, the time period was between two executions of men whose name was synonymous with that ideology's stated aim.

Yeah, that's not what your OP states, it says 'someone', not two men. Changes the meaning entirely.

[NickKIELCEPoland]

Yeah, that's not what your OP states, it says 'someone', not two men. Changes the meaning entirely.
Indeed.

[dilettante]

Any chance we could get the answer to this?

[Michael]

Any chance we could get the answer to this?

Indeed, I second this request. And I will promise not to critique it if I don't like it. :)

[NickKIELCEPoland]

Communism - its stated aim was victory for the people, but its heyday began with the excution of Tsar Nicholas II (Nicholas means 'victory for the people')

Its heyday ended around 1989 with mostly peaceful revolutions, but with the execution of Nicolae Ceausescu, Nicolae being a variation of Nicholas.
;)

[Michael]

Communism - its stated aim was victory for the people, but its heyday began with the excution of Tsar Nicholas II (Nicholas means 'victory for the people')

Its heyday ended around 1989 with mostly peaceful revolutions, but with the execution of Nicolae Ceausescu, Nicolae being a variation of Nicholas.
;)

Okay, that does work as a solution.

(Though it is a bit stretched since the People's Republic of China, Castro's Cuba and the crackpots of N.Korea still officially pay lip service to the virtues of communism - not to mention the Orwellian usage of 'victory' for the people of Eastern Europe or USSR). ;)